Bob was tasked with finding the limit of the function $$ f(x) = \frac{\sqrt{\sin^2 x}}{x} $$ at the point $x=0$.
At first, he simplified the function: $$ \lim_{x\rightarrow 0} \frac{\sqrt{\sin^2 x}}{x} = \lim_{x\rightarrow 0} \frac{\sin x}{x} $$ Then, he applied l’Hospital’s rule and evaluated the limit: $$ \lim_{x\rightarrow 0} \frac{\sin x}{x} = \lim_{x\rightarrow 0} \frac{\cos x}{1} = 1 $$ Did Bob calculate the limit correctly? Explain.
No. It holds $\sqrt{\sin^2x }=|\sin x |$, and the function $\frac{|\sin x |}{x}$ has no limit at the point $x=0$.
No. It isn’t possible to apply l’Hospital’s rule to $\lim_{x\rightarrow 0} \frac{\sin x}{x}$.
Yes. The limit is calculated correctly.
No. It should be $\lim_{x\rightarrow 0} \frac{\sin x}{x} =0$.
The function $f(x) = \frac{\sqrt{\sin^2 x}}{x} =\frac{|\sinx |}{x}$ has no limit at the point $x=0$ since the limit from the left equals $−1$ while the limit from the right equals $+1$. The one-sided limits of the function $f$ can be clearly seen from its graph shown in the figure below.
